Download 2014

BANNER ID B 0 0 __ __ __ __ __ __
1
INTRODUCTORY BIOCHEMISTRY
BIOL0280
First Midterm Examination
February 25, 2014
Enter Legible BANNER ID: B 0 0 __ __ __ __ __ __
DO NOT WRITE YOUR NAME ON THIS EXAM
Make sure that your Banner ID is on every page. This is the only way we have of matching you with
your exam after grading it. Please work independently. Read each question carefully before answering.
Unless otherwise indicated, there is only one correct answer for each multiple-choice question. Points
are indicated by the question within brackets []. There are no calculators or other electronic devices
needed or allowed on this exam. All hats must be removed during the exam. Exams will be scanned
electronically before being returned.
Page 2 total __________/8
Page 3 total __________/10
Page 4 total __________/14
Page 5 total __________/11
Page 6 total __________/14
Page 7 total __________/12
Page 8 total __________/8
Page 9 total __________/13
Page 10 total _________/10
Exam total __________/100
BANNER ID B 0 0 __ __ __ __ __ __
2
1. [4 points] Given that pK1 = 2; pK2 = 9; pKR = 11, Draw the ionization states and calculate the pI of
lysine. Write the PI on the line below
3points for correct ionization state structures, 1 point for correct pI
PI=9+11/2=10
pI = 10
3 points for drawing and 1 point for PI
2. [4 points] Draw the structure of Tyr-Ser-Asp in the ionic form that predominates at pH 7. Draw all
atoms (including hydrogens) as well as any charges necessary.
O
H 3N
CH C
CH2
O
H
N
CH C
O
H
N
CH2
OH
CH C
O
CH2
O
C
O
OH
2 points for correct structure (side chains and amide bond linkages), 1 point for putting correct
amino acids at the n terminus and c terminus, 1 point for getting the ionization states correct
including all side chains and the n and c terminus
BANNER ID B 0 0 __ __ __ __ __ __
3
3. [2 points] The amino acid substitution of Val for Glu in Hemoglobin S results in aggregation of the
protein because of ___________ interactions between molecules.
A) covalent
B) disulfide
C) hydrogen bonding
D) hydrophobic
E) ionic
Circle the correct answer.
4. [2 points] In the sequencing of peptides by tandem mass spectrometry, peptide molecules are
fragmented into smaller molecules to determine the amino acid sequence of the peptide. The types
of bonds broken to generate MSMS tandem mass spectra are
A) carbon-carbon bonds within the amino acid side chains
B) disulfide bonds
C) amide bonds
D) bonds between alpha carbons and amino groups in the polypeptide backbone
E) bonds between alpha carbons and carbonyl groups in the polypeptide backbone
Circle the correct answer.
5. [2 points]
A
B
Leu-Ala-Phe Tyr-Lys-Met
Which one of the above tripeptides:
C
Asp-Pro-Glu
D
Gly-Pro-Arg
E
Asp-Trp-Tyr
__C__(a) is most negatively charged at pH 7?
__A__(b) contains the largest number of nonpolar R groups?
6. [2 points] An α helix would be destabilized most by:
A) interactions between neighboring Asp and Arg residues.
B) interactions between two adjacent hydrophobic Val residues.
C) the presence of an Arg residue near the carboxyl terminus of the α helix.
D) the presence of two Lys residues near the amino terminus of the α helix.
E) the presence of a Glu residue near the amino terminus of the α helix.
Circle the correct answer.
7. [2 points] The major reason that antiparallel β-stranded protein structures are more stable than
parallel β-stranded structures is that the latter:
A) are in a slightly less extended configuration than antiparallel strands.
B) do not have as many disulfide crosslinks between adjacent strands.
C) do not stack in sheets as well as antiparallel strands.
D) have fewer lateral hydrogen bonds than antiparallel strands.
E) have weaker hydrogen bonds laterally between adjacent strands.
Circle the correct answer.
BANNER ID B 0 0 __ __ __ __ __ __
4
8. [2 points] Experiments on denaturation and renaturation in the enzyme ribonuclease (RNase) by Dr.
Anfinsen have shown that:
A) folding of denatured RNase into the native, active conformation, requires the input of energy in
the form of heat.
B) native ribonuclease does not have a unique secondary and tertiary structure.
C) the completely unfolded enzyme, with all —S—S— bonds broken, is still enzymatically active.
D) the enzyme, dissolved in water, is thermodynamically stable relative to the mixture of amino
acids whose residues are contained in RNase.
E) the primary sequence of RNase is sufficient to determine its specific secondary and tertiary
structure.
Circle the correct answer.
9. [2 points] Which one phrase best describes the hydrogen bonds in a β sheet?
A) they occur mainly near the amino and carboxyl termini of the strands.
B) they are perpendicular to the plane of the sheet.
C) they occur mainly between atoms of the R groups.
D) they occur only between some of the amino acids of each strand.
E) they occur mainly between atoms of adjacent strands in a sheet.
Circle the correct answer.
10. [6 points] Supply the missing words in each of the following 3 sentences. Choose from increase,
decrease, or remain unchanged.
Increasing the concentration of HCO3– in a solution of hemoglobin (without changing the pH)
causes the affinity for O2 to _____decrease_.
Increasing the pH of a solution of hemoglobin causes the affinity for O2 to _______ increase __.
Removing 2,3 BPG from a solution of adult hemoglobin causes the affinity for O2 to _ increase _.
14. [4 points] Name two examples of proteins with atypical protein structures discussed in lecture 2 and
explain briefly the key characteristics of these atypical protein structures.
2 points for naming them. Myosin, keratin or collagen
2 points for explaining the concept of coiled-coils.
1 point for either of the following concepts:
- myosin and keratin contain two right-handed helices twisted in a left-handed superhelix
- amphiphilic helices
- collagen contains three left-handed helices in a right-handed superhelix
- rope-like structures
-3.5 residues per turn myosin keratin
-collagen contains Gly at center of triple helix
-collagen contains Pro hydroxyproline
BANNER ID B 0 0 __ __ __ __ __ __
5
12. [2 points] Enzymes catalyze the rapid formation of product by:
A) shifting the position of the equilibrium between substrate and product in favor of product.
B) lowering ΔG´o of the reaction.
C) raising the free energy of the substrate.
D) raising ΔG‡
E) none of the above.
Circle the correct answer.
13. [5 points] Write out the simple chemical reaction that describes the mechanism for enzyme action
used as a model by Michaelis and Menten. Indicate the rate constants for all steps in this reaction.
List the assumptions used by Michaelis and Menten to derive a rate equation for this reaction.
Finally write down the Michaelis Menten equation.
The chemical reaction (with rate constants) is:
2 points
k1 k2
k3
E+S
ES
EP
E+P
k-1 k-2
k-3
Assumptions:
One assumption is that significant product has not accumulated, so that the rate of the reaction
depends exclusively on the breakdown of ES and is not influenced by the reverse reaction; that is, k-2
can be ignored and V0 = k2 [ES]. This condition is possible only if early reaction times are
measured; the velocity, therefore, is an initial velocity. A second assumption is that the rate of ES
formation equals the rate of ES breakdown; in other words, the reaction is at a steady state. A third
assumption is [S] >> [Et], so that total [S], which equals free substrate and enzyme-bound substrate,
is essentially equal to [S].
2 points for assumptions (1 each)
Michaelis Menten equation:
V0=Vmax[S]/(Km+[S])
1 point
11. [4 points] In lecture 3 we discussed the T and R states describing the quaternary structure of
hemoglobin. Describe briefly the two principal models for the cooperative binding of ligands to
proteins such as hemoglobin with multiple ligand binding sites. Be sure to name both models and
provide an explanation of the differences between the two.
Ans: In the concerted model, binding of a ligand to one site on one subunit results in an
allosteric effect that converts all of the remaining subunits to the high-affinity conformation.
As a result, all of the subunits are either in the low- or high-affinity conformation. In the
sequential model, each subunit is changed individually to the high affinity conformation. As a
result, there are many possible combinations of low- and high-affinity subunits.
BANNER ID B 0 0 __ __ __ __ __ __
6
15. [2 points] An enzyme follows Michaelis-Menten kinetics. Indicate (with an "x") which of the kinetic
parameters at the left would be altered by the following factors. Give only one answer for each.
Km
Vmax
Neither
Both
Factor
X
competitive inhibitor
X
noncompetitive inhibitor
16. [2 points] Which amino acids in chymotrypsin are found in the active site and are participants in
substrate cleavage?
A) his, ser, asp
B) his, ser, glu
C) asp, lys
D) lys, arg
E) his, ser, arg
Circle the correct answer.
17. [2 points] How is specificity determined by chymotrypsin?
A) interaction of the substrate with the oxyanion hole
B) binding of the N-terminus amino acid at the active site
C) covalent binding of a his residue to the substrate
D) conformational change upon binding of substrate
E) binding of the proper amino acid into a deep pocket on the enzyme
Circle the correct answer.
18. [2 points] In the following oligopeptide, indicate by drawing vertical line(s) where chymotrypsin
would hydrolyze peptide bond(s). (You cannot receive credit if your vertical line(s) is/are drawn
through a letter) :
D---A---V---T---W--|-C---L---A---F--|-T---K---S---G
19. [2 points] Catalytic antibodies are most likely to be active if they are made by using an antigen that:
A) resembles the substrate.
B) is composed of an equimolar mixture of substrate and product.
C) resembles the enzyme-substrate complex.
D) resembles a transition state intermediate.
E) is an allosteric modulator.
Circle the correct answer.
20. [4 points] The Michaelis-Menten constant, Km, is actually a summary of three rate constants. Write
down the equation to calculate Km based on these rate constants. Describe a method to determine Km
graphically indicating precisely how Km is measured.
Km= (k2 + k–1)/k1
Km can be determined graphically on a plot of V0 vs. [S] by finding the [S] at which V0 = 1/2
Vmax. More conveniently, on a double-reciprocal plot, the x-axis intercept = –1/ Km.
2 points for Km equation, 2 points for Km from V vs. S plot, or 2 points for double reciprocal
plot x intercept of -1/Km
BANNER ID B 0 0 __ __ __ __ __ __
7
21. [2 points] How is chymotrypsinogen converted to chymotrypsin?
A) A protein kinase-catalyzed phosphorylation converts chymotrypsinogen to chymotrypsin.
B) An increase in Ca2+ concentration promotes the conversion.
C) Proteolysis of chymotrypsinogen forms chymotrypsin.
D) Chymotrypsinogen dimers bind an allosteric modulator, cAMP, causing dissociation into active
chymotrypsin monomers.
E) Two inactive chymotrypsinogen dimers pair to form an active chymotrypsin tetramer.
Circle the correct answer.
22. [2 points] The enzyme ATCase is regulated by CTP, which binds to the T-state of ATCase. CTP is
a:
A) heterotropic positive allosteric regulator.
B) heterotropic negative allosteric regulator.
C) cofactor.
D) competitive inhibitor.
E) homotropic negative allosteric regulator.
Circle the correct answer.
23. [2 points] Circle the fatty acid in each pair that has the lower melting temperature.
A) 18:1Δ9
18:2Δ9,12
B) 16:0
18:0
24. [2 points] Vitamin A (retinol), vitamin K (phylloquinone), testosterone, and cortisol are all members
of a single family of lipids. Draw the structure of the basic building block that is used to assemble all of
these molecules.
25. [4 points] The bacterium E. coli can grow at 20 °C or at 40 °C. At which growth temperature would
you expect the membrane phospholipids to have a lower ratio of unsaturated to saturated fatty
acids, and why?
At 40 °C, the membranes of E. coli will contain a lower ratio of unsaturated to saturated fatty
acids than at 20 °C. The cell regulates fatty acid composition to achieve the same fluidity in its
membranes, regardless of growth temperature. Saturated fatty acids counterbalance the
fluidizing effect of high temperature by providing additional hydrophobic interactions.
BANNER ID B 0 0 __ __ __ __ __ __
8
26. [2 points] Glucose transport through intestinal epithelial cells involves two types of transporters.
Glucose is first transported into the intestinal epithelial cell by _________ and then it is transported out
by _________.
A) secondary active transport / primary active transport
B) facilitated diffusion / secondary active transport
C) primary active transport / secondary active transport
D) secondary active transport / facilitated diffusion
E) secondary active transport / secondary active transport
Circle the correct answer.
27. [2 points] In one catalytic cycle, the Na+/K+ ATPase transporter transports:
A) 2 Na+ out, 3 K+ in, and converts 1 ATP to ADP + Pi.
B) 3 Na+ out, 2 K+ in, and converts 1 ATP to ADP + Pi.
C) 3 Na+ in, 2 K+ out, and converts 1 ATP to ADP + Pi.
D) 1 Na+ out, 1 K+ in, and converts 1 ATP to ADP + Pi.
E) 2 Na+ out, 3 K+ in, and converts 1 ADP + Pi to ATP.
Circle the correct answer.
28. [2 points] Which one of the following chemical modifications (I-IV) would result in the conversion
of D-fructofuranose from a reducing carbohydrate to a non-reducing carbohydrate?
I
A) II
B) I
C) IV
D) III
E) None of the above
Circle the correct answer.
CH3
O
H3C
IV
O
e
HO
a
a
bH
c
b
H
OH
O
d
H
III
OH
OH
II
c
f
f
H2C
H
OH
H
H2C
e
OH
CH3
O
d
O
H
CH3
29. [2 points] Label each of the following as an aldose, a ketose, or neither.
H
H
H
H
C
OH
O
C
C
O
H
C
C
OH
HO
C
H
H
C
OH
OH
HO
C
H
H
HO
C
H
H
H
(a)
(b)
___Ketose__
__Aldose___
H
H
HO
C
H
H
C
OH
C
O
C
O
H
C
OH
H
C
H
H
H
H
(c)
(d)
(e)
__neither__
___aldose_
BANNER ID B 0 0 __ __ __ __ __ __
9
30. [4 points] Indicate in the blanks provided the match between the molecules and their biological
roles.
(a) peptidoglycan
____d____ homopolysaccharide of glucose in animals
(b) starch
____b____ homopolysaccharide of glucose in plants
(c) chitin
____c____ exoskeleton of lobsters
(d) glycogen
____a____ structural component of bacterial cell walls
33. [3 points] Indicate the oxidation state of each carbon in the fructose molecule shown below:
Oxidation State
C1
_____5______
C2
_____2______
C3
_____4______
C4
_____4______
C5
_____4______
C6
_____5______
31. [2 points] What factor contributes most significantly to the thermodynamic stability of the
monosaccharide glucose relative to other monosaccharide six carbon D-aldoses?
Most thermodynamically stable form of glucose is the form with all non-hydrogen substituents
in the equatorial position. No other 6 carbon D-aldose is fully equatorial
32. [4 points] (a) Describe one biological advantage of storing glucose units in branched polymers
(glycogen, amylopectin) rather than in linear polymers.
The enzymes that act on these polymers to mobilize glucose for metabolism act only on their
nonreducing ends. With extensive branching, there are more such ends for enzymatic attack
than would be present in the same quantity of glucose stored in a linear polymer. In effect,
branched polymers increase the substrate concentration for these enzymes.
(b) What is the biological advantage to an organism that stores its carbohydrate reserves as starch or
glycogen rather than as an equivalent amount of free glucose?
The polymers are essentially insoluble and contribute little to the osmolarity of the cell,
thereby avoiding the influx of water that would occur with the glucose in solution. They also
make the uptake of glucose energetically more feasible than it would be with free glucose in the
cell. Full credit for recognition of concentration difference or full credit for concept of less
reducing ends
BANNER ID B 0 0 __ __ __ __ __ __
10
34. [2 points] Standard reduction potentials are tabulated as: Oxidized form → Reduced form +
electrons.
A) True
B) False
Catabolic pathways diverge, anabolic pathways converge.
A) True
B) False
Circle the correct answer.
36. [2 points] One molecule of NADH is produced per glucose in the conversion of glucose to lactate
A) True
B) False
Circle the correct answer.
Triose phosphate isomerase decreases ΔG for the step in glycolysis catalyzed by aldolase through
rapid removal of dihydroxyacetone phosphate.
A) True
B) False
Circle the correct answer
37. [2 points] The pentose phosphate pathway can alternatively be called the pentose phosphate cycle
because _______ is a net product of the pathway that can be recycled.
A) phosphate
B) NADP+
C) carbon dioxide
D) ribose 5-phosphate
E) UDP-glucose
Circle the correct answer.
38. [2 points] The conversion of pyruvate to ethanol also causes the ______.
A) Oxidation of NADH
B) Production of ADP
C) Release of CO2
D) Generation of an ion gradient across mitochondrial membranes
E) Both A and C
Circle the correct answer.
39. [2 points] ATP is both a substrate and a negative effector (inhibitor) for the allosteric, glycolytic
enzyme phosphofructokinase (PFK). Which of the following answers best explains this observation?
A) ATP has equal affinity for the active site and the regulatory (allosteric) site of PFK
B) ATP only binds the regulatory site of PFK at low concentrations
C) ATP only binds the regulatory site of PFK at high concentrations
D) ATP and AMP are both negative effectors of PFK
Circle the correct answer.